$.ajax post empty data to server with jsonp - php

this below code could'nt send data to other server. i want to send "aaa-bbb-ccc" with $.ajax. but after post back userCode thats post empty data from $_POST. sorry for my english
jquery code :
<script type="text/javascript">
$(function(){
$.ajax({
url: "http://www.site.com/index.php",
type: "POST",
dataType: "jsonp",
data: {userCode: "aaa-bbb-ccc"}
}).done(function(data){
alert(data.message);
});
});
</script>
server index.php :
<?php
include_once ('./AFactory.class.php');
$database= new AFactory;
$db=new AFactory();
$link=$db->getDBO();
if ( $_POST['userCode'] == '')
{
$data['success']=false;
$data['message']='ERROR ...';
}
else {
$query=array('id'=>NULL,'userCode'=>$_POST['userCode']);
$sql=$db->insertQuery('`alachiq_takhmis`.`users`',$query);
if ( mysql_query($sql) )
{
$data['success']=true;
$data['message']=$_POST['userCode'];
}
else
{
$data['success']=false;
$data['message']=$_POST['userCode'];
}
}
echo $_GET['callback'] . '('. json_encode($data) . ')';
?>
post back:
({"success":false,"message":'ERROR ...'})
whats my code problem?

JSONP works by injecting a <script> element with a src attribute into a document.
That can only ever make a GET request.

$.ajax({
url: "http://www.site.com/index.php",
type: "GET",
dataType: "jsonp",
data: {userCode: "aaa-bbb-ccc"}
});

Related

Ajax Post to PHP returning empty array

I need to pass a json object from JS to PHP, and it will pass, but the result is an empty array.
Ajax request in 'adopt.php':
var info = JSON.stringify(filteredArray);
$.ajax({
type: 'POST',
url: 'ajax.php',
data: {'info': info},
success: function(data){
console.log(data);
}
});
ajax.php code:
if(isset($_POST['info'])){
$_SESSION['array'] = $_POST['info'];
}
back in adopt.php, later:
if(isset($_SESSION['array'])){
$arr = $_SESSION['array'];
echo "console.log('information: ' + $arr);";
}
in both of the console.logs, it returns an empty object. Does anybody know what could be causing this? (i've tried just passing the json without stringifying it, but it throws a jquery error whenever i do this.)
Try below code i think you miss return ajax response
adopt.php
<script>
var info = JSON.stringify(filteredArray);
$.ajax({
type: 'POST',
url: 'ajax.php',
data: {info: info},
success: function(data){
console.log(data);
}
});
</script>
ajax.php
if (isset($_POST['info'])) {
$_SESSION['array'] = $_POST['info'];
echo json_encode(["result" => "success"]);
}
To get the response data from PHP, you need to echo your data to return it to the browser.
In your ajax.php:
if (isset($_POST['info'])) {
$_SESSION['array'] = $_POST['info'];
echo json_encode(['result' => $_SESSION['array']]);
}
Your ajax.php is not returning any data.To get data at the time of success you need to echo the data you want to display on success of your ajax.

How to send json to php via ajax?

I have a form that collect user info. I encode those info into JSON and send to php to be sent to mysql db via AJAX. Below is the script I placed before </body>.
The problem now is, the result is not being alerted as it supposed to be. SO I believe ajax request was not made properly? Can anyone help on this please?Thanks.
<script>
$(document).ready(function() {
$("#submit").click(function() {
var param2 = <?php echo $param = json_encode($_POST); ?>;
if (param2 && typeof param2 !== 'undefined')
{
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: param2,
cache: false,
success: function(result) {
alert(result);
}
});
}
});
});
</script>
ajaxsubmit.php
<?php
$phpArray = json_decode($param2);
print_r($phpArray);
?>
You'll need to add quotes surrounding your JSON string.
var param2 = '<?php echo $param = json_encode($_POST); ?>';
As far as I am able to understand, you are doing it all wrong.
Suppose you have a form which id is "someForm"
Then
$(document).ready(function () {
$("#submit").click(function () {
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: $('#someForm').serialize(),
cache: false,
success: function (result) {
alert(result);
}
});
}
});
});
In PHP, you will have something like this
$str = "first=myName&arr[]=foo+bar&arr[]=baz";
to decode
parse_str($str, $output);
echo $output['first']; // myName
For JSON Output
echo json_encode($output);
If you are returning JSON as a ajax response then firstly you have define the data type of the response in AJAX.
try it.
<script>
$(document).ready(function(){
$("#submit").click(function(){
var param2 = <?php echo $param = json_encode($_POST); ?>
if( param2 && typeof param2 !== 'undefined' )
{
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: dataString,
cache: false,
dataType: "json",
success: function(result){
alert(result);
}
});}
});
});
</script>
It's just really simple!
$(document).ready(function () {
var jsonData = {
"data" : {"name" : "Randika",
"age" : 26,
"gender" : "male"
}
};
$("#getButton").on('click',function(){
console.log("Retrieve JSON");
$.ajax({
url : "http://your/API/Endpoint/URL",
type: "POST",
datatype: 'json',
data: jsonData,
success: function(data) {
console.log(data); // any response returned from the server.
}
});
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="submit" value="POST JSON" id="getButton">
For your further readings and reference please follow the links bellow:
Link 1 - jQuery official doc
Link 2 - Various types of POSTs and AJAX uses.
In my example, code snippet PHP server side should be something like as follows:
<?php
$data = $_POST["data"];
echo json_encode($data); // To print JSON Data in PHP, sent from client side we need to **json_encode()** it.
// When we are going to use the JSON sent from client side as PHP Variables (arrays and integers, and strings) we need to **json_decode()** it
if($data != null) {
$data = json_decode($data);
$name = $data["name"];
$age = $data["age"];
$gender = $data["gender"];
// here you can use the JSON Data sent from the client side, name, age and gender.
}
?>
Again a code snippet more related to your question.
// May be your following line is what doing the wrong thing
var param2 = <?php echo $param = json_encode($_POST); ?>
// so let's see if param2 have the reall json encoded data which you expected by printing it into the console and also as a comment via PHP.
console.log("param2 "+param2);
<?php echo "// ".$param; ?>
After some research on the google , I found the answer which alerts the result in JSON!
Thanks for everyone for your time and effort!
<script>
$("document").ready(function(){
$(".form").submit(function(){
var data = {
"action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "response.php", //Relative or absolute path to response.php file
data: data,
success: function(data) {
$(".the-return").html(
"<br />JSON: " + data["json"]
);
alert("Form submitted successfully.\nReturned json: " + data["json"]);
}
});
return false;
});
});
</script>
response.php
<?php
if (is_ajax()) {
if (isset($_POST["action"]) && !empty($_POST["action"])) { //Checks if action value exists
$action = $_POST["action"];
switch($action) { //Switch case for value of action
case "test": test_function(); break;
}
}
}
//Function to check if the request is an AJAX request
function is_ajax() {
return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest';
}
function test_function(){
$return = $_POST;
echo json_encode($return);
}
?>
Here's the reference link : http://labs.jonsuh.com/jquery-ajax-php-json/

Using Jquery AJAX to POST variable in PHP

I believe I have the right syntax but am missing something important. Been searching on here for a while but can't figure out why the POST variable is not being detected. Basically my .ajax is firing because my test statement has been changing due to the value but some reason can't receive variable via $_POST (i.e. my echo in php echo that it is not firing) Also the native file and php that I am sending it to are the same file blankFormTemplate.php but don't think that should be an issue.
$(document).ready(function()
{
var $selectedContexts = [];
$('.allContextField').change(function(){
//alert($(this).val());
hideField = $(this).val();
$('#'+hideField).remove();
$.ajax
({
type: "POST",
url: "blankFormTemplate.php",
data: addedContext=hideField,
cache: false,
success: function(addedContext)
{
$('#test').html($('#test').html()+hideField);
}
});
});
});
in my PHP blankFormTemplate.php:
<?php
if(isset($_POST['addedContext']))
{
echo 'hello';
}
else
{
echo 'why';
}
?>
Any help would be greatly appreciated.
Thanks,
Your javascript
$(document).ready(function()
{
$('.allContextField').change(function(){
hideField = $(this).val();
$('#'+hideField).remove();
});
if (hideField.trim()) {
$.ajax
({
type: "POST",
url: "blankFormTemplate.php",
data: (addedContext:hideField},
cache: false,
success: function(msg)
{
$('#test').html(msg);
}
});
}
});
Your PHP
<?php
if(isset($_POST['addedContext']))
{
echo 'hello';
}
else
{
echo 'why';
}
?>

Jquery ajax POST response is null

I have a js script that does an ajax request and posts the data to a php script, this script with then echo something back depending if it works or not.
here is the JS
$(document).ready(function(){
var post_data = [];
$('.trade_window').load('signals.php?action=init');
setInterval(function(){
post_data = [ {market_number:1, name:$('.trade_window .market_name_1').text().trim()},
{market_number:2, name:$('.trade_window .market_name_2').text().trim()}];
$.ajax({
url: 'signals.php',
type: 'POST',
contentType: 'application/json; charset=utf-8',
data:{markets:post_data},
dataType: "json",
success: function(response){
console.log("Response was " + response);
},
failure: function(result){
console.log("FAILED");
console.log(result);
}
});
}, 6000);
});
here is the php:
if(isset($_POST["json"]))
{
$json = json_decode($_POST["json"]);
if(!empty($json))
{
echo "IT WORKED!!!!";
}
else
echo "NOT POSTED";
}
So basically, i thought the response in the `success: function(response)' method would be populated with either "IT WORKED!!!" or "NOT POSTED" depending on the if statement in the php. Now everything seem to work because the js script manages to go into the success statement but prints this to the console:
Response was null
I need to be able to get the return from the server in order to update the screen.
Any ideas what I'm doing wrong?
Try:
if(isset($_POST["markets"]))
{
$json = json_decode($_POST["markets"]);
if(!empty($json))
{
echo "IT WORKED!!!!";
}
else
echo "NOT POSTED";
}
use this in your php file
if(isset($_POST["markets"]))
{
}
instead of
if(isset($_POST["json"]))
{
.
.
.
.
}
Obiously the if(isset($_POST["json"])) statement is not invoked, so neither of both echos is executed.
The fact that the function specified in .ajax success is invoked, only tells you that the http connection to the url was successful, it does not indicate successful processing of the data.
You are using "success:" wrong.
Try this instead.
$.post("signals.php", { markets: post_data }).done(function(data) {
/* This will return either "IT WORKED!!!!" or "NOT POSTED" */
alert("The response is: " + data);
});
Also have a look at the jQuery documentation.
http://api.jquery.com/jQuery.post/
Look, You send data in market variable not in json. Please change on single.php code by this.
$json_data = array();
if(isset($_POST["markets"]))
{
// $json = json_decode($_POST["markets"]);
$json = ($_POST["markets"]);
if(!empty($json))
echo "IT WORKED!!!!";
else
echo "NOT POSTED";
}
And change on your ajax function
$(document).ready(function(){
var post_data = [];
$('.trade_window').load('signals.php?action=init');
setInterval(function(){
post_data = [ {market_number:1, name:$('.trade_window .market_name_1').text().trim()},
{market_number:2, name:$('.trade_window .market_name_2').text().trim()}];
$.ajax({
url: 'signals.php',
type: 'post',
// contentType: 'application/json; charset=utf-8',
data:{markets:post_data},
dataType: "json",
success: function(response){
console.log("Response was " + response);
},
failure: function(result){
console.log("FAILED");
console.log(result);
}
});
},6000);
});
You have to you change you $.ajax call with
//below post_data array require quotes for keys like 'market_number' and update with your required data
post_data = [ {'market_number':1, 'name':'name1'},
{'market_number':2, 'name':'name2'}];
//console.log(post_data);
$.ajax({
url: "yourfile.php",
type:'post',
async: true,
data:{'markets':post_data},
dataType:'json',
success: function(data){
console.log(data);
},
});
and you php file will be
<?php
if(isset($_POST['markets']))
{
echo "It worked!!!";
}
else
{
echo "It doesn't worked!!!";
}
//if you want to work with json then below will help you
//$data = json_encode($_POST['markets']);
//print_r($data);
?>
in your php file check the $_POST:
echo(json_encode($_POST));
which will tell if your data has been posted or not and the data structure in $_POST.
I have used the following code to covert the posted data to associative array:
$post_data = json_decode(json_encode($_POST), true);

return php variable to jquery ajax

I have an ajax function in jquery calling a php file to perform some operation on my database, but the result may vary. I want to output a different message whether it succeeded or not
i have this :
echo '<button id="remove_dir" onclick="removed('.$dir_id.')">remove directory</button>';
<script type="text/javascript">
function removed(did){
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did},
success: function(rmd){
if(rmd==0)
alert("deleted");
else
alert("not empty");
window.location.reload(true);
}
});
}
</script>
and this
<?php
require('bdd_connect.php');
require('functions/file_operation.php');
if(isset($_POST['dir_id'])){
$rmd=remove_dir($_POST['dir_id'],$bdd);
}
?>
my question is, how to return $rmd so in the $.ajax, i can alert the correct message ?
thank you for your answers
PHP
<?php
require('bdd_connect.php');
require('functions/file_operation.php');
if (isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo $rmd;
}
?>
JS
function removed(did){
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did}
}).done(function(rmd) {
if (rmd===0) {
alert("deleted");
}else{
alert("not empty");
window.location.reload(true);
}
});
}
i advice to use json or :
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo $rmd;
}
You need your php file to send something back, then you need the ajax call on the original page to behave based on the response.
php:
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo "{'rmd':$rmd}";
}
which will output one of two things: {"rmd": 0} or {"rmd": 1}
We can simulate this return on jsBin
Then use jquery to get the value and do something based on the response in our callback:
$.ajax({
type: "POST",
dataType: 'json',
url: "http://jsbin.com/iwokag/3",
success: function(data){
alert('rmd = ' + data.rmd)
}
});
View the code, then watch it run.
Only I didn't send any data here, my example page always returns the same response.
Just try echoing $rmd in your ajax file, and then watching the console (try console.log(rmd) in your ajax response block)
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did},
success: function(rmd){
console.log(rmd);
}
});
You can then act accordingly based on the response
Try echo the $rmd out in the php code, as an return to the ajax.
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
//if $rmd = 1 alert('directory not empty');
//if $rmd = 0 alert('directory deleted');
echo $rmd;
}
Your "rmd" in success: function(rmd) should receive the callabck.

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